Problem: Different dealers may sell the same car for different prices. The sale prices for a particular car are normally distributed with a mean and standard deviation of $26$ thousand dollars and $2$ thousand dollars, respectively. Suppose we select one of these cars at random. Let $X=$ the sale price (in thousands of dollars) for the selected car. Find $P(26<X<30)$. You may round your answer to two decimal places.
Representing probability with area Since we know the distribution of scores is normally distributed, the probability $P(26<X<30)$ can be found by calculating the shaded area between $X=26$ and $X=30$ in the corresponding normal distribution: $20$ $22$ $24$ $26$ $28$ $30$ $32$ $ \mu_X = 26$ $ \sigma_X = 2$ $ P(26<X<30)$ Calculating shaded area We can use the "normalcdf" function on most graphing calculators to find the shaded area: $\begin{aligned} &\text{normalcdf:} \\\\ &\text{lower bound: } 26 \\\\ &\text{upper bound: } 30 \\\\ &\mu=26 \\\\ &\sigma=2 \end{aligned}$ Output: $\approx0.4772$ [Why do we use normalcdf instead of normalpdf?] Answer $P(26<X<30)\approx0.48$ [Is there another way?]